Problem: $w_1=6[\cos(19^\circ)+i\sin(19^\circ)]$ $w_2=2[\cos(8^\circ)+i\sin(8^\circ)]$ Express the quotient, $\dfrac{w_1}{w_2}$, in polar form. Angle measure should be in degrees. No need to use the degree symbol. $\dfrac{w_1}{w_2}= $
Solution: Background For any two complex numbers $z_1$ and $z_2$ (whose radii are $r_1$ and $r_2$ and angles are $\theta_1$ and $\theta_2$ ): The radius of $\dfrac{z_1}{z_2}$ is the quotient of the original radii, $\dfrac{r_1}{r_2}$. The angle of $\dfrac{z_1}{z_2}$ is the difference of the original angles, $\theta_1 - \theta_2$. In other words, suppose the polar forms of $z_1$ and $z_2$ are as follows, $z_1 = r_1[\cos(\theta_1) + {i}\sin(\theta_1)]$ $z_2 = r_2[\cos(\theta_2) + {i}\sin(\theta_2)]$, then the polar form of their quotient is: $\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + {i}\sin(\theta_1 - \theta_2)]$. [How do we get this?] Finding the radius of $\dfrac{w_1}{w_2}$ $w_1=6[\cos(19^\circ)+i\sin(19^\circ)]$ $w_2=2[\cos(8^\circ)+i\sin(8^\circ)]$ Here, $r_1=6$ and $r_2=2$. Therefore, the radius of $\dfrac{w_1}{w_2}$ is $\dfrac{r_1}{r_2}=3$. Finding the angle of $\dfrac{w_1}{w_2}$ $w_1=6[\cos(19^\circ)+i\sin(19^\circ)]$ $w_2=2[\cos(8^\circ)+i\sin(8^\circ)]$ Here, $\theta_1=19^\circ$ and $\theta_2=8^\circ$. Therefore, the angle of $\dfrac{w_1}{w_2}$ is $\theta_1-\theta_2=19^\circ-8^\circ=11^\circ$ Summary We found that the radius of $\dfrac{w_1}{w_2}$ is $3$ and its angle is $11^\circ$. Therefore, $\dfrac{w_1}{w_2}=3\left(\cos(11^\circ)+i\sin(11^\circ)\right)$